Note that the cube \(H^3\)  is planar so that \(CCC^3\)  is planar.
Figure REF  shows a 1-planar drawing of \(H^4\) ,
which can be transformed into a RAC drawing, and an IC-planar drawing \(CCC^4\) , so
that these graphs are 1-planar and RAC [1]}. Every cube-connected cycle
is 1-bend RAC [2]}, that is \(\sigma _1(CCC^d)\)  is RAC. We can
summarize:
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